®
Integration by Parts
13.4
Introduction
Integration by Parts is a technique for integrating products of functions. In this Section you will learn
to recognise when it is appropriate to use the technique and have the opportunity to practise using
it for finding both definite and indefinite integrals.
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• understand what is meant by definite and
indefinite integrals
Prerequisites
Before starting this Section you should . . .
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• be able to use a table of integrals
• be able to differentiate and integrate a range
of common functions
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• decide when it is appropriate to use the
method known as integration by parts
Learning Outcomes
• apply the formula for integration by parts to
definite and indefinite integrals
On completion you should be able to . . .
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HELM (2008):
Section 13.4: Integration by Parts
• perform integration by parts repeatedly if
appropriate
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33
1. Indefinite integration
The technique known as integration by parts is used to integrate a product of two functions, such
as in these two examples:
Z
Z 1
2x
x3 e−2x dx
(i)
e sin 3x dx
(ii)
0
Note that in the first example, the integrand is the product of the functions e2x and sin 3x, and in
the second example the integrand is the product of the functions x3 and e−2x . Note also that we
can change the order of the terms in the product if we wish and write
Z
Z 1
2x
e−2x x3 dx
(i)
(sin 3x) e dx
(ii)
0
What you must never do is integrate each term in the product separately and then multiply - the
integral of a product is not the product of the separate integrals. However, it is often possible to
find integrals involving products using the method of integration by parts - you can think of this as
a product rule for integrals.
The integration by parts formula states:
Key Point 5
Integration by Parts for Indefinite Integrals
For indefinite integrals, given functions f (x) and g(x):
Z
Z
Z Z
df
f · g dx = f · g dx −
· gdx dx
dx
Alternatively, given functions u and v:
Z
Z
dv
du
u dx = u.v − v dx
dx
dx
Study the formula carefully and note the following observations. Firstly, to apply the formula we must
df
be able to differentiate the function f to find
, and we must be able to integrate the function, g.
dx
Secondly the formula replaces one integral, the one on the left, with a different integral, that on the
far right. The intention is that the latter, whilst it may look more complicated in the formula above,
is simpler to evaluate. Consider the following Example:
34
HELM (2008):
Workbook 13: Integration
®
Example 15
Find the integral of the product of x with sin x; that is, find
R
x sin x dx.
Solution
Compare the required integral with the formula for integration by parts: we choose
f =x
and
It follows that
df
=1
dx
g = sin x
Z
and
Z
g dx =
sin x dx = − cos x
(When integrating g there is no need to worry about a constant of integration. When you become
confident with the method, you may like to think about why this is the case.)
Applying the formula we obtain
Z
Z
Z Z
df
x sin x dx = f · g dx −
· gdx dx
dx
Z
= x(− cos x) − 1(− cos x) dx
Z
= −x cos x + cos x dx = −x cos x + sin x + c
Task
Z
Find
(5x + 1) cos 2x dx.
df
and
Let f = 5x + 1 and g = cos 2x. Now calculate
dx
Z
g dx:
Your solution
Answer
Z
df
1
= 5 and cos 2x dx = sin 2x.
dx
2
Substitute these results into the formula for integration by parts and complete the Task:
Your solution
HELM (2008):
Section 13.4: Integration by Parts
35
Answer
1
(5x + 1)( sin 2x) −
2
Z
1
1
5
5( sin 2x)dx = (5x + 1) sin 2x + cos 2x + c
2
2
4
Sometimes it is necessary to apply the formula more than once, as the next Example shows.
Example
16
Z
2x2 e−x dx
Find
Solution
df
We let f = 2x and g = e . Then
= 4x and
dx
2
−x
Z
gdx = −e−x
Using the formula for integration by parts we find
Z
Z
Z
2 −x
2
−x
−x
2 −x
2x e dx = 2x (−e ) − 4x(−e )dx = −2x e + 4xe−x dx
R
We now need to find 4xe−x dx using integration by parts again. We get
Z
Z
−x
−x
4xe dx = 4x(−e ) − 4(−e−x )dx
Z
−x
= −4xe + 4e−x dx = −4xe−x − 4e−x
Altogether we have
Z
2x2 e−x dx = −2x2 e−x − 4xe−x − 4e−x + c = −2e−x (x2 + 2x + 2) + c
Exercises
In some questions
below it will be necessary
to apply integration
by parts more than once.
Z
Z
Z
3t
1. Find (a) x sin(2x)dx,
(b) te dt,
(c) x cos x dx.
Z
2. Find (x + 3) sin x dx.
Z
3. By writing ln x as 1 × ln x find ln x dx.
Z
Z
Z
−1
4. Find (a) tan x dx, (b) −7x cos 3x dx, (c) 5x2 e3x dx,
Z
Z
5. Find (a) x cos kx dx, where k is a constant
(b) z 2 cos kz dz, where k is a constant.
Z
Z
−st
(b) Find t2 e−st dt where s is a constant.
6. Find (a) te dt where s is a constant,
36
HELM (2008):
Workbook 13: Integration
®
Answers
1
1
1. (a) sin 2x − x cos 2x + c,
4
2
2. −(x + 3) cos x + sin x + c.
1
1
(b) e3t ( t − ) + c,
3
9
(c) cos x + x sin x + c
3. x ln x − x + c.
1
7
7
5
ln(x2 + 1) + c, (b) − cos 3x − x sin 3x + c, (c) e3x (9x2 − 6x + 2) + c,
2
9
3
27
2z cos kz z 2 sin kz 2 sin kz
cos kx x sin kx
+
c,
(b)
−
5. (a)
+
+
+ c.
k2
k
k2
k
k3
−e−st (st + 1)
−e−st (s2 t2 + 2st + 2)
6. (a)
+
c,
(b)
+ c.
s2
s3
4. (a) x tan−1 x −
2. Definite integration
When dealing with definite integrals the relevant formula is as follows:
Key Point 6
Integration by Parts for Definite Integrals
For definite integrals, given functions f (x) and g(x):
Z
b Z b Z b
Z
df
f · g dx = f · g dx −
· gdx dx
dx
a
a
a
b Z b
Z b
dv
du
u dx = uv −
Alternatively, given functions u and v:
v dx
dx
dx
a
a
a
Example
17
Z
2
xex dx.
Find
0
Solution
Z
df
We let f = x and g = e . Then
= 1 and g dx = ex . Using integration by parts we obtain
dx
2 Z 2
2
Z 2
x
x
x
2
xe dx = xe −
1.e dx = 2e − ex = 2e2 −[e2 −1] = e2 +1
(or 8.389 to 3 d.p.)
x
0
0
0
0
Sometimes it is necessary to apply the formula more than once as the next Example shows.
HELM (2008):
Section 13.4: Integration by Parts
37
Example 18
Z
2 x
2
x2 ex dx.
Find the definite integral of x e from 0 to 2; that is, find
0
Solution
2
x
We let f = x and g = e . Then
Z
2
2 x
2 x
2
0
Z
and
2
x
2
Z
g dx = ex . Using integration by parts:
2
2xe dx = 4e − 2
−
x e dx = x e
0
Z
df
= 2x
dx
xex dx
0
0
The remainingZ integral must be integrated by parts also but we have just done this in the example
2
x2 ex dx = 4e2 − 2[e2 + 1] = 2e2 − 2 = 12.778
(3 d.p.)
above. So
0
Task
Z
π/4
(4 − 3x) sin x dx.
Find
0
What are your choices for f, g?
Your solution
Answer
Take f = 4 − 3x and g = sin x.
Now complete the integral:
Your solution
Z π/4
(4 − 3x) sin x dx =
0
Answer
Z
π/4
(4 − 3x) sin x dx =
π/4
Z
(4 − 3x)(− cos x)
−3
0
=
(4 − 3x)(− cos x)
π/4
−3
0
cos x dx
0
0
π/4
π/4
sin x
0
= 0.716 to 3 d.p.
38
HELM (2008):
Workbook 13: Integration
®
Exercises
Z
1. Evaluate the following: (a)
Z 2
(x + 2) sin x dx
2. Find
1
Z 1
(x2 − 3x + 1)ex dx
3. Find
1
Z
x cos 2x dx,
(b)
0
π/2
Z
1
x sin 2x dx, (c)
0
te2t dt
−1
0
Answers
1. (a) 0.1006,
(b) π/4 = 0.7854,
(c) 1.9488.
2. 3.3533.
3. −0.5634.
HELM (2008):
Section 13.4: Integration by Parts
39